Ways of arranging people around a round table such that persons of same nationality are separated.
$begingroup$
There are 3 Americans, 2 British, 1 Chinese, 1 Dutch and 1 Egyptian. They are to be arranged around a circular table so that persons of same nationality are separated.
(I found a similar thread on this site but wasn't able to get my answer since I am unable to form the case when ONLY two Americans are separated.
Ways of arranging of different nationality persons at a round table )
combinatorics permutations
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
$endgroup$
add a comment |
$begingroup$
There are 3 Americans, 2 British, 1 Chinese, 1 Dutch and 1 Egyptian. They are to be arranged around a circular table so that persons of same nationality are separated.
(I found a similar thread on this site but wasn't able to get my answer since I am unable to form the case when ONLY two Americans are separated.
Ways of arranging of different nationality persons at a round table )
combinatorics permutations
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
$endgroup$
add a comment |
$begingroup$
There are 3 Americans, 2 British, 1 Chinese, 1 Dutch and 1 Egyptian. They are to be arranged around a circular table so that persons of same nationality are separated.
(I found a similar thread on this site but wasn't able to get my answer since I am unable to form the case when ONLY two Americans are separated.
Ways of arranging of different nationality persons at a round table )
combinatorics permutations
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
$endgroup$
There are 3 Americans, 2 British, 1 Chinese, 1 Dutch and 1 Egyptian. They are to be arranged around a circular table so that persons of same nationality are separated.
(I found a similar thread on this site but wasn't able to get my answer since I am unable to form the case when ONLY two Americans are separated.
Ways of arranging of different nationality persons at a round table )
combinatorics permutations
combinatorics permutations
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
edited Dec 27 '18 at 19:47
Arpit Kumar
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
asked Dec 27 '18 at 8:18
Arpit KumarArpit Kumar
113
113
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
I provide a startup for you.
Place an American.
Then $5$ spots are left for the other two Americans in the sense that they cannot be placed next to the one that has a seat already.
$binom52$ ways to find $2$ out of $5$ but - because they cannot sit next to each other $4$ possibilities fall out. Next to that as human beings they are distinguishable so order matters here.
This gives $2left(binom52-4right)=12$ possibilities.
These possibilities can be split in two specific configurations (make a picture to get view on this) of $6$.
Now start placing the $2$ British in both configurations.
I leave the rest to you.
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
$endgroup$
$begingroup$
Can't this be done by inclusion exclusion principle? We can take $E_1$ - Americans are sitting together and $E_2$ - British are sitting together
$endgroup$
– Sauhard Sharma
Dec 27 '18 at 10:32
$begingroup$
Oh Wow, finally got it! Thanks for the help. Luckily both the configurations of Americans yielded same number of purmutations.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 19:58
$begingroup$
Also found an alternate solution to it by myself. I placed One British, Chinese, Dutch and Egyptian first. Then placed the remaining British either adjacent to the first British which forced me to place an American between the two or placed the second British alternate to the first one. Then I simply used gap method to fill in the Americans.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:02
$begingroup$
I still wonder though, just as @Sauhard said, if we could use inclusion exclusion principle on it(though it would be quite long) . I tried it but as I said I failed to account for ONLY two Americans sitting together. I wonder if you could help me with that.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:05
$begingroup$
@ArpitKumar I encountered (so understand) the same problem with that and for me it was a reason not to choose for PIE.
$endgroup$
– drhab
Dec 28 '18 at 8:42
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
I provide a startup for you.
Place an American.
Then $5$ spots are left for the other two Americans in the sense that they cannot be placed next to the one that has a seat already.
$binom52$ ways to find $2$ out of $5$ but - because they cannot sit next to each other $4$ possibilities fall out. Next to that as human beings they are distinguishable so order matters here.
This gives $2left(binom52-4right)=12$ possibilities.
These possibilities can be split in two specific configurations (make a picture to get view on this) of $6$.
Now start placing the $2$ British in both configurations.
I leave the rest to you.
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
$endgroup$
$begingroup$
Can't this be done by inclusion exclusion principle? We can take $E_1$ - Americans are sitting together and $E_2$ - British are sitting together
$endgroup$
– Sauhard Sharma
Dec 27 '18 at 10:32
$begingroup$
Oh Wow, finally got it! Thanks for the help. Luckily both the configurations of Americans yielded same number of purmutations.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 19:58
$begingroup$
Also found an alternate solution to it by myself. I placed One British, Chinese, Dutch and Egyptian first. Then placed the remaining British either adjacent to the first British which forced me to place an American between the two or placed the second British alternate to the first one. Then I simply used gap method to fill in the Americans.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:02
$begingroup$
I still wonder though, just as @Sauhard said, if we could use inclusion exclusion principle on it(though it would be quite long) . I tried it but as I said I failed to account for ONLY two Americans sitting together. I wonder if you could help me with that.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:05
$begingroup$
@ArpitKumar I encountered (so understand) the same problem with that and for me it was a reason not to choose for PIE.
$endgroup$
– drhab
Dec 28 '18 at 8:42
|
show 1 more comment
$begingroup$
I provide a startup for you.
Place an American.
Then $5$ spots are left for the other two Americans in the sense that they cannot be placed next to the one that has a seat already.
$binom52$ ways to find $2$ out of $5$ but - because they cannot sit next to each other $4$ possibilities fall out. Next to that as human beings they are distinguishable so order matters here.
This gives $2left(binom52-4right)=12$ possibilities.
These possibilities can be split in two specific configurations (make a picture to get view on this) of $6$.
Now start placing the $2$ British in both configurations.
I leave the rest to you.
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
$endgroup$
$begingroup$
Can't this be done by inclusion exclusion principle? We can take $E_1$ - Americans are sitting together and $E_2$ - British are sitting together
$endgroup$
– Sauhard Sharma
Dec 27 '18 at 10:32
$begingroup$
Oh Wow, finally got it! Thanks for the help. Luckily both the configurations of Americans yielded same number of purmutations.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 19:58
$begingroup$
Also found an alternate solution to it by myself. I placed One British, Chinese, Dutch and Egyptian first. Then placed the remaining British either adjacent to the first British which forced me to place an American between the two or placed the second British alternate to the first one. Then I simply used gap method to fill in the Americans.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:02
$begingroup$
I still wonder though, just as @Sauhard said, if we could use inclusion exclusion principle on it(though it would be quite long) . I tried it but as I said I failed to account for ONLY two Americans sitting together. I wonder if you could help me with that.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:05
$begingroup$
@ArpitKumar I encountered (so understand) the same problem with that and for me it was a reason not to choose for PIE.
$endgroup$
– drhab
Dec 28 '18 at 8:42
|
show 1 more comment
$begingroup$
I provide a startup for you.
Place an American.
Then $5$ spots are left for the other two Americans in the sense that they cannot be placed next to the one that has a seat already.
$binom52$ ways to find $2$ out of $5$ but - because they cannot sit next to each other $4$ possibilities fall out. Next to that as human beings they are distinguishable so order matters here.
This gives $2left(binom52-4right)=12$ possibilities.
These possibilities can be split in two specific configurations (make a picture to get view on this) of $6$.
Now start placing the $2$ British in both configurations.
I leave the rest to you.
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
$endgroup$
I provide a startup for you.
Place an American.
Then $5$ spots are left for the other two Americans in the sense that they cannot be placed next to the one that has a seat already.
$binom52$ ways to find $2$ out of $5$ but - because they cannot sit next to each other $4$ possibilities fall out. Next to that as human beings they are distinguishable so order matters here.
This gives $2left(binom52-4right)=12$ possibilities.
These possibilities can be split in two specific configurations (make a picture to get view on this) of $6$.
Now start placing the $2$ British in both configurations.
I leave the rest to you.
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
answered Dec 27 '18 at 9:37
drhabdrhab
102k545136
102k545136
$begingroup$
Can't this be done by inclusion exclusion principle? We can take $E_1$ - Americans are sitting together and $E_2$ - British are sitting together
$endgroup$
– Sauhard Sharma
Dec 27 '18 at 10:32
$begingroup$
Oh Wow, finally got it! Thanks for the help. Luckily both the configurations of Americans yielded same number of purmutations.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 19:58
$begingroup$
Also found an alternate solution to it by myself. I placed One British, Chinese, Dutch and Egyptian first. Then placed the remaining British either adjacent to the first British which forced me to place an American between the two or placed the second British alternate to the first one. Then I simply used gap method to fill in the Americans.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:02
$begingroup$
I still wonder though, just as @Sauhard said, if we could use inclusion exclusion principle on it(though it would be quite long) . I tried it but as I said I failed to account for ONLY two Americans sitting together. I wonder if you could help me with that.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:05
$begingroup$
@ArpitKumar I encountered (so understand) the same problem with that and for me it was a reason not to choose for PIE.
$endgroup$
– drhab
Dec 28 '18 at 8:42
|
show 1 more comment
$begingroup$
Can't this be done by inclusion exclusion principle? We can take $E_1$ - Americans are sitting together and $E_2$ - British are sitting together
$endgroup$
– Sauhard Sharma
Dec 27 '18 at 10:32
$begingroup$
Oh Wow, finally got it! Thanks for the help. Luckily both the configurations of Americans yielded same number of purmutations.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 19:58
$begingroup$
Also found an alternate solution to it by myself. I placed One British, Chinese, Dutch and Egyptian first. Then placed the remaining British either adjacent to the first British which forced me to place an American between the two or placed the second British alternate to the first one. Then I simply used gap method to fill in the Americans.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:02
$begingroup$
I still wonder though, just as @Sauhard said, if we could use inclusion exclusion principle on it(though it would be quite long) . I tried it but as I said I failed to account for ONLY two Americans sitting together. I wonder if you could help me with that.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:05
$begingroup$
@ArpitKumar I encountered (so understand) the same problem with that and for me it was a reason not to choose for PIE.
$endgroup$
– drhab
Dec 28 '18 at 8:42
$begingroup$
Can't this be done by inclusion exclusion principle? We can take $E_1$ - Americans are sitting together and $E_2$ - British are sitting together
$endgroup$
– Sauhard Sharma
Dec 27 '18 at 10:32
$begingroup$
Can't this be done by inclusion exclusion principle? We can take $E_1$ - Americans are sitting together and $E_2$ - British are sitting together
$endgroup$
– Sauhard Sharma
Dec 27 '18 at 10:32
$begingroup$
Oh Wow, finally got it! Thanks for the help. Luckily both the configurations of Americans yielded same number of purmutations.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 19:58
$begingroup$
Oh Wow, finally got it! Thanks for the help. Luckily both the configurations of Americans yielded same number of purmutations.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 19:58
$begingroup$
Also found an alternate solution to it by myself. I placed One British, Chinese, Dutch and Egyptian first. Then placed the remaining British either adjacent to the first British which forced me to place an American between the two or placed the second British alternate to the first one. Then I simply used gap method to fill in the Americans.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:02
$begingroup$
Also found an alternate solution to it by myself. I placed One British, Chinese, Dutch and Egyptian first. Then placed the remaining British either adjacent to the first British which forced me to place an American between the two or placed the second British alternate to the first one. Then I simply used gap method to fill in the Americans.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:02
$begingroup$
I still wonder though, just as @Sauhard said, if we could use inclusion exclusion principle on it(though it would be quite long) . I tried it but as I said I failed to account for ONLY two Americans sitting together. I wonder if you could help me with that.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:05
$begingroup$
I still wonder though, just as @Sauhard said, if we could use inclusion exclusion principle on it(though it would be quite long) . I tried it but as I said I failed to account for ONLY two Americans sitting together. I wonder if you could help me with that.
$endgroup$
– Arpit Kumar
Dec 27 '18 at 20:05
$begingroup$
@ArpitKumar I encountered (so understand) the same problem with that and for me it was a reason not to choose for PIE.
$endgroup$
– drhab
Dec 28 '18 at 8:42
$begingroup$
@ArpitKumar I encountered (so understand) the same problem with that and for me it was a reason not to choose for PIE.
$endgroup$
– drhab
Dec 28 '18 at 8:42
|
show 1 more comment
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